Chapter 4, Problem 56P

To determine

Part(a)To Determine:

The distance the block would go up an inclined plane when given an initial velocity.

Answer to Problem 56P

Solution:

The distance the block goes up the block is 2.8 m.

Explanation of Solution

A block is given an initial speed slides up a distance s along the length of the incline which is at an angle $\theta$ and comes to rest after traveling a distance s up the plane.

The free body diagram for the block is shown in the diagram below.

The weight mg of the block acts vertically downwards. The normal force $F_N$ acts perpendicular to the incline. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the incline.

The block experiences a retarding force due to the component of weight $mg\sin \theta$, as it moves up and is decelerated at the rate a.

    $F=-mg\sin \theta$……(1)

By Newton’s second law,

    $F=ma$……(2)

From equations (1)and (2),

    $\begin{array}{c}ma=-mg\sin \theta \\ a=-g\sin \theta \end{array}$……(3)

Given:

The initial speed of the block $v_0=4.5\text{ m/s}$

The angle of the incline $\theta =22.0°$

The speed of the block at the distance s is $v=0\text{ m/s}$

Formula used:

    $v^2=v_0^2-2sg\sin \theta$

Calculation:

The distance the block travels up the incline before coming to rest is calculated by substituting the given values of v,$v_0$, $\theta$ and $\left(9.8{\text{ m/s}}^2\right)$ for g in the equation $v^2=v_0^2-2sg\sin \theta$ and solve for s.

$\begin{array}{c}{v}^2=v_0^2-2sg\sin \theta \\ {\left(0\text{ m/s}\right)}^2={\left(4.5\text{ m/s}\right)}^2-2s\left(9.8{\text{ m/s}}^2\right)\left(\sin 22°\right)\\ s=\frac{{\left(4.5\text{ m/s}\right)}^2}{2\left(9.8{\text{ m/s}}^2\right)\left(\sin 22°\right)}\\ =2.76\text{ m}\\ =\text{2}\text{.8 m}\end{array}$

To determine

Part(b)To Determine:

The time taken by the block to return to the starting point.

Answer to Problem 56P

Solution:

The time taken by the block to return to the starting point is 2.4 s.

Explanation of Solution

To determine the distance traveled up the plane using the equation of motion,

    $v^2=v_0^2+2as$

Use equation (3)in the above equation.

    $v^2=v_0^2-2sg\sin \theta$…….(4)

To determine the time taken by the block to return to the starting point, use the equation of motion

    $s=v_{0}t+\frac{1}{2}a{t}^2$

Use equation (3)in the above equation.

    $s=v_{0}t-\frac{1}{2}\left(g\sin \theta \right)t^2$……(5)

Given:

The initial speed of the block $v_0=4.5\text{ m/s}$

The angle of the incline $\theta =22.0°$

The speed of the block at the distance s is $v=0\text{ m/s}$

Formula used:

    $s=v_{0}t-\frac{1}{2}\left(g\sin \theta \right)t^2$

Calculation:

When the block reaches the starting point, its net displacement s_net is zero. Use 0 for s and the given values of v,$v_0$ , $\theta$ and $\left(9.8{\text{ m/s}}^2\right)$ for g in the equation $s=v_{0}t-\frac{1}{2}\left(g\sin \theta \right)t^2$ to solve for t.

$\begin{array}{c}{s}_{net}=v_{0}t-\frac{1}{2}\left(g\sin \theta \right)t^2\\ \left(0\text{ m}\right)=\left(4.5\text{ m/s}\right)t-\frac{1}{2}\left(9.8{\text{ m/s}}^2\right)\left(\sin 22°\right)t^2\end{array}$

The equation has two roots, $t=0$ and

    $\begin{array}{c}t=\frac{2\left(4.5\text{ m/s}\right)}{\left(9.8{\text{ m/s}}^2\right)\left(\sin 22°\right)}\\ =2.45\text{ s}\end{array}$

Since $t=0$ refers to the time at which the block is set into motion, the time taken to reach the starting point is 2.4 s (correct to 2 sf).